First Order RL Circuits

Time constant, current rise and fall.

Darshan N
Updated: 19 March 2026
4 min read

A first order RL circuit consists of a resistor and an inductor. Like the RC circuit, it responds to sudden changes in input with a transient exponential behavior, but the energy is stored in the magnetic field of the inductor rather than the electric field of a capacitor. RL circuits are critical in understanding inductive loads, relay coils, transformers, and power electronics switching.

First Order RL Circuit OverviewVoltageSource VsResistorRInductorLi(t)Drop: i·REMF: L·di/dtTime constant:τ = L/Ri(t) = (Vs/R)(1 - e^(-t/τ)) for step input
Figure 1: RL circuit topology with current direction and voltage drop labels for inductor and resistor

Core Concept Explanation

When a DC voltage source is suddenly connected to an RL circuit, the inductor opposes the sudden change in current through it. This opposition is caused by the back EMF generated by the inductor according to Faraday's law: V = L·di/dt. At t = 0+, the inductor acts like an open circuit because it resists any instantaneous change in current. As time progresses, the inductor's opposition reduces and the current gradually rises toward its final value of Vs/R.

The time constant of the RL circuit is τ = L/R. A larger inductance means more energy needs to be stored before current reaches its final value, making the transient slower. A larger resistance dissipates energy faster, making the transient quicker. This is the opposite intuition from RC circuits where larger R makes the transient slower.

The source-free response of an RL circuit occurs when a steady current is flowing through the inductor and the source is suddenly removed (or the circuit is opened). The inductor now acts as a source, maintaining current flow through the resistor, and the current decays exponentially to zero. This is why inductor current at t = 0+ equals inductor current at t = 0−: current through an inductor cannot change instantaneously.

Mathematical Expression

Applying KVL to the RL series circuit: Vs = i·R + L·di/dt. This is a first-order linear differential equation. The homogeneous solution gives the natural response and the particular solution gives the forced response. The complete response is: i(t) = (Vs/R) + (I0 - Vs/R)·e^(-t/τ), where I0 is the initial current through the inductor and τ = L/R.

For a step input with zero initial current: i(t) = (Vs/R)·(1 - e^(-t/τ)). The current starts at 0, rises exponentially, and reaches Vs/R at steady state. Voltage across the inductor: VL(t) = Vs·e^(-t/τ), which starts at Vs and decays to 0 as the inductor behaves as a short circuit to DC at steady state.

Practical Understanding

When a relay coil (inductive load) is switched off, the current does not stop instantly. The inductor drives a large voltage spike (sometimes called inductive kickback) as it tries to maintain current flow. This voltage can damage transistors and other switching devices. A freewheeling diode is placed across the inductor to provide a safe path for this current, which is a direct application of RL transient behavior.

In power electronics, the RL time constant determines how fast a circuit can switch. A motor winding with high inductance and low resistance will have a large τ, meaning it cannot rapidly change its current. Engineers must account for this when designing motor drives and converters.

Example
Given:
R = 50 Ω, L = 200 mH, Vs = 10 V, Initial i(0) = 0 A

Why this formula applies:
Step voltage applied to RL with zero initial current. Current rises exponentially.

Formula:
i(t) = (Vs/R)(1 - e^(-t/τ))   where τ = L/R

Substitution:
τ = 200×10⁻³ / 50 = 4 ms
Final steady-state current = Vs/R = 10/50 = 0.2 A
i(τ) = 0.2 × (1 - e^(-1)) = 0.2 × 0.632

Calculation:
i(τ) = 0.1264 A ≈ 126.4 mA
VL at t=0+ = Vs = 10 V
VL at t=τ: VL = 10·e^(-1) = 10 × 0.368 = 3.68 V

Final Answer: At t = 4 ms, current is 126.4 mA (63.2% of 200 mA). Inductor voltage drops from 10 V to 3.68 V at t = τ.
Exam Tip: For GATE, note that iL(0+) = iL(0−) — inductor current cannot change instantaneously. Also, at steady state, inductor is a short circuit (VL = 0) in DC circuits. Voltage across inductor is maximum at t = 0+ and decays to zero — opposite of capacitor voltage in RC circuit.

RL Transient Waveforms

RL Circuit: Current Rise and Inductor Voltage ResponseCurrent i(t) — Step InputTime →Vs/Rτ63.2%Inductor Voltage VL(t)Time →Vsτ36.8%SummaryCurrent rises: i(t) = (Vs/R)(1 - e^(-t/τ)) VL decays: VL(t) = Vs·e^(-t/τ)As i → Vs/R (short circuit), VL → 0 (no back EMF needed) τ = L/R
Figure 2: Current build-up and inductor voltage decay in a first order RL circuit. Note complementary behavior at t = τ.
  • At t = 0+, inductor acts as open circuit: full source voltage appears across it. Current starts at zero.
  • Current rises exponentially following i(t) = (Vs/R)(1 - e^(-t/τ)), approaching Vs/R asymptotically.
  • Inductor voltage VL decays from Vs to 0. At steady state, inductor is a short circuit for DC.
  • During source-free (natural) response, i(t) = I0·e^(-t/τ) — current decays exponentially through the resistor.
  • Switching off an inductive circuit causes a large voltage spike. Protective diodes (freewheeling diodes) are used to prevent device damage in practical switching circuits.

Quick Revision

  • Time constant τ = L/R in seconds. Larger L or smaller R means slower transient.
  • Step response current: i(t) = (Vs/R)(1 - e^(-t/τ)). At t = τ: 63.2% of final value.
  • Source-free current: i(t) = I0·e^(-t/τ). Decays exponentially from initial value.
  • iL(0+) = iL(0−): inductor current cannot change instantaneously — key initial condition rule.
  • At steady state, inductor = short circuit (VL = 0). At t = 0+, inductor = open circuit (max VL).
  • Exam trap: Current rises in RL while voltage rises in RC at step input. Current and voltage roles are swapped between the two circuits.
  • Inductive kickback voltage can be very large if R is small. Always account for initial stored energy when computing initial conditions.

RL Circuit Analysis

Sharpen your analysis of current transients and time constants in first-order RL circuits.

Question 1 of 3

Q1.A series RL circuit has R = 50 ohm and L = 500 mH. A DC voltage of 100 V is applied at t = 0 with zero initial current. What is the steady-state current and the time constant?